3.16.5 \(\int \frac {e^{\frac {2}{100 x+10 e^{1+e^{e^x}} x}} (-10-1500 x-15 e^{2+2 e^{e^x}} x+e^{1+e^{e^x}} (-1-300 x-e^{e^x+x} x))}{500 x^5+100 e^{1+e^{e^x}} x^5+5 e^{2+2 e^{e^x}} x^5} \, dx\)

Optimal. Leaf size=26 \[ \frac {e^{\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}}}{x^3} \]

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Rubi [F]  time = 18.78, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{\frac {2}{100 x+10 e^{1+e^{e^x}} x}} \left (-10-1500 x-15 e^{2+2 e^{e^x}} x+e^{1+e^{e^x}} \left (-1-300 x-e^{e^x+x} x\right )\right )}{500 x^5+100 e^{1+e^{e^x}} x^5+5 e^{2+2 e^{e^x}} x^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2/(100*x + 10*E^(1 + E^E^x)*x))*(-10 - 1500*x - 15*E^(2 + 2*E^E^x)*x + E^(1 + E^E^x)*(-1 - 300*x - E^(
E^x + x)*x)))/(500*x^5 + 100*E^(1 + E^E^x)*x^5 + 5*E^(2 + 2*E^E^x)*x^5),x]

[Out]

-1/5*Defer[Int][E^(1 + E^E^x + 1/(5*(10 + E^(1 + E^E^x))*x))/((10 + E^(1 + E^E^x))^2*x^5), x] - 2*Defer[Int][E
^(1/(5*(10 + E^(1 + E^E^x))*x))/((10 + E^(1 + E^E^x))^2*x^5), x] - 60*Defer[Int][E^(1 + E^E^x + 1/(5*(10 + E^(
1 + E^E^x))*x))/((10 + E^(1 + E^E^x))^2*x^4), x] - 3*Defer[Int][E^(2*(1 + E^E^x) + 1/(5*(10 + E^(1 + E^E^x))*x
))/((10 + E^(1 + E^E^x))^2*x^4), x] - 300*Defer[Int][E^(1/(5*(10 + E^(1 + E^E^x))*x))/((10 + E^(1 + E^E^x))^2*
x^4), x] - Defer[Int][E^(1 + E^E^x + E^x + 1/(5*(10 + E^(1 + E^E^x))*x) + x)/((10 + E^(1 + E^E^x))^2*x^4), x]/
5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}} \left (-10-1500 x-15 e^{2+2 e^{e^x}} x+e^{1+e^{e^x}} \left (-1-300 x-e^{e^x+x} x\right )\right )}{5 \left (10+e^{1+e^{e^x}}\right )^2 x^5} \, dx\\ &=\frac {1}{5} \int \frac {e^{\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}} \left (-10-1500 x-15 e^{2+2 e^{e^x}} x+e^{1+e^{e^x}} \left (-1-300 x-e^{e^x+x} x\right )\right )}{\left (10+e^{1+e^{e^x}}\right )^2 x^5} \, dx\\ &=\frac {1}{5} \int \left (-\frac {e^{1+e^{e^x}+\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}}}{\left (10+e^{1+e^{e^x}}\right )^2 x^5}-\frac {10 e^{\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}}}{\left (10+e^{1+e^{e^x}}\right )^2 x^5}-\frac {300 e^{1+e^{e^x}+\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}}}{\left (10+e^{1+e^{e^x}}\right )^2 x^4}-\frac {15 e^{2 \left (1+e^{e^x}\right )+\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}}}{\left (10+e^{1+e^{e^x}}\right )^2 x^4}-\frac {1500 e^{\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}}}{\left (10+e^{1+e^{e^x}}\right )^2 x^4}-\frac {e^{1+e^{e^x}+e^x+\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}+x}}{\left (10+e^{1+e^{e^x}}\right )^2 x^4}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{1+e^{e^x}+\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}}}{\left (10+e^{1+e^{e^x}}\right )^2 x^5} \, dx\right )-\frac {1}{5} \int \frac {e^{1+e^{e^x}+e^x+\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}+x}}{\left (10+e^{1+e^{e^x}}\right )^2 x^4} \, dx-2 \int \frac {e^{\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}}}{\left (10+e^{1+e^{e^x}}\right )^2 x^5} \, dx-3 \int \frac {e^{2 \left (1+e^{e^x}\right )+\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}}}{\left (10+e^{1+e^{e^x}}\right )^2 x^4} \, dx-60 \int \frac {e^{1+e^{e^x}+\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}}}{\left (10+e^{1+e^{e^x}}\right )^2 x^4} \, dx-300 \int \frac {e^{\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}}}{\left (10+e^{1+e^{e^x}}\right )^2 x^4} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.34, size = 26, normalized size = 1.00 \begin {gather*} \frac {e^{\frac {1}{5 \left (10+e^{1+e^{e^x}}\right ) x}}}{x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2/(100*x + 10*E^(1 + E^E^x)*x))*(-10 - 1500*x - 15*E^(2 + 2*E^E^x)*x + E^(1 + E^E^x)*(-1 - 300*x
- E^(E^x + x)*x)))/(500*x^5 + 100*E^(1 + E^E^x)*x^5 + 5*E^(2 + 2*E^E^x)*x^5),x]

[Out]

E^(1/(5*(10 + E^(1 + E^E^x))*x))/x^3

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fricas [A]  time = 0.77, size = 29, normalized size = 1.12 \begin {gather*} \frac {e^{\left (\frac {1}{5 \, {\left (x e^{\left ({\left (e^{\left (x + e^{x}\right )} + e^{x}\right )} e^{\left (-x\right )}\right )} + 10 \, x\right )}}\right )}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x*exp(exp(exp(x))+1)^2+(-x*exp(x)*exp(exp(x))-300*x-1)*exp(exp(exp(x))+1)-1500*x-10)*exp(1/(10*
x*exp(exp(exp(x))+1)+100*x))^2/(5*x^5*exp(exp(exp(x))+1)^2+100*x^5*exp(exp(exp(x))+1)+500*x^5),x, algorithm="f
ricas")

[Out]

e^(1/5/(x*e^((e^(x + e^x) + e^x)*e^(-x)) + 10*x))/x^3

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x*exp(exp(exp(x))+1)^2+(-x*exp(x)*exp(exp(x))-300*x-1)*exp(exp(exp(x))+1)-1500*x-10)*exp(1/(10*
x*exp(exp(exp(x))+1)+100*x))^2/(5*x^5*exp(exp(exp(x))+1)^2+100*x^5*exp(exp(exp(x))+1)+500*x^5),x, algorithm="g
iac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{4687500000000,[0,8,6,6,15,8]%%%}+%%%{31250000000,[0,8,6,
6,14,8]%%%}

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maple [A]  time = 0.08, size = 21, normalized size = 0.81

method	result	size
risch	\(\frac {{\mathrm e}^{\frac {1}{5 \left ({\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}+1}+10\right ) x}}}{x^{3}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-15*x*exp(exp(exp(x))+1)^2+(-x*exp(x)*exp(exp(x))-300*x-1)*exp(exp(exp(x))+1)-1500*x-10)*exp(1/(10*x*exp(
exp(exp(x))+1)+100*x))^2/(5*x^5*exp(exp(exp(x))+1)^2+100*x^5*exp(exp(exp(x))+1)+500*x^5),x,method=_RETURNVERBO
SE)

[Out]

1/x^3*exp(1/5/(exp(exp(exp(x))+1)+10)/x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{5} \, \int \frac {{\left (15 \, x e^{\left (2 \, e^{\left (e^{x}\right )} + 2\right )} + {\left (x e^{\left (x + e^{x}\right )} + 300 \, x + 1\right )} e^{\left (e^{\left (e^{x}\right )} + 1\right )} + 1500 \, x + 10\right )} e^{\left (\frac {1}{5 \, {\left (x e^{\left (e^{\left (e^{x}\right )} + 1\right )} + 10 \, x\right )}}\right )}}{x^{5} e^{\left (2 \, e^{\left (e^{x}\right )} + 2\right )} + 20 \, x^{5} e^{\left (e^{\left (e^{x}\right )} + 1\right )} + 100 \, x^{5}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x*exp(exp(exp(x))+1)^2+(-x*exp(x)*exp(exp(x))-300*x-1)*exp(exp(exp(x))+1)-1500*x-10)*exp(1/(10*
x*exp(exp(exp(x))+1)+100*x))^2/(5*x^5*exp(exp(exp(x))+1)^2+100*x^5*exp(exp(exp(x))+1)+500*x^5),x, algorithm="m
axima")

[Out]

-1/5*integrate((15*x*e^(2*e^(e^x) + 2) + (x*e^(x + e^x) + 300*x + 1)*e^(e^(e^x) + 1) + 1500*x + 10)*e^(1/5/(x*
e^(e^(e^x) + 1) + 10*x))/(x^5*e^(2*e^(e^x) + 2) + 20*x^5*e^(e^(e^x) + 1) + 100*x^5), x)

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mupad [B]  time = 1.19, size = 20, normalized size = 0.77 \begin {gather*} \frac {{\mathrm {e}}^{\frac {1}{50\,x+5\,x\,\mathrm {e}\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}}}}}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2/(100*x + 10*x*exp(exp(exp(x)) + 1)))*(1500*x + 15*x*exp(2*exp(exp(x)) + 2) + exp(exp(exp(x)) + 1)*
(300*x + x*exp(exp(x))*exp(x) + 1) + 10))/(100*x^5*exp(exp(exp(x)) + 1) + 5*x^5*exp(2*exp(exp(x)) + 2) + 500*x
^5),x)

[Out]

exp(1/(50*x + 5*x*exp(1)*exp(exp(exp(x)))))/x^3

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sympy [A]  time = 1.93, size = 20, normalized size = 0.77 \begin {gather*} \frac {e^{\frac {2}{10 x e^{e^{e^{x}} + 1} + 100 x}}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x*exp(exp(exp(x))+1)**2+(-x*exp(x)*exp(exp(x))-300*x-1)*exp(exp(exp(x))+1)-1500*x-10)*exp(1/(10
*x*exp(exp(exp(x))+1)+100*x))**2/(5*x**5*exp(exp(exp(x))+1)**2+100*x**5*exp(exp(exp(x))+1)+500*x**5),x)

[Out]

exp(2/(10*x*exp(exp(exp(x)) + 1) + 100*x))/x**3

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